Week 2

10.30 A. hypotheses. Ho: p (red) p (yellow) > 0 Ha: p (red) p (yellow) is not greater than 0 B. 1st ending One tail w/ of import =1% circumstantial value is z = 2.326 C. examine proportions and z statistic. p hold (red) = 20/153,348 ; p hat (yellow) = 4/135,035 For Pooling: p relegateber = (20+4) / (153348+135035) = 0.00002.08 z = [(20/153,348) - (4/135,035)] / sqrt [(0.00002.08)(0.999979) / 153348 + (0.00002.08)*0.999979/135035)] = 2.960988... D. 2nd Decision sightedness as tryout statistic is more than particular value, reject Ho E. p - value and take in it. p - value = P(2.960988 < z < 10) = 0.00153 0.1533% run across that the test has intemperateer evidence against Ho. F. If statistically pregnant, do you think the release is life-size enough to be important? If so, to whom, and why? Yes it is statistically crucial; yellow paint has reduced accidents. G. Is the normality hypothesis fulfill? Explain. Yes: p1n1 > 5, q1n 1 > 5 ; p2n2 > 5, p2n2 > 5 10.44 A. Hypotheses. Ho: p(inactive) - P(active) = 0 Ha: p(inactive) - p(active) > 0 p-hat(inactive) = 97 / 2081 = 0.0466 p-hat(active) = 57 / 2325 = 0.0245 Pool the data you get p - bar = (97 + 57) / (2081 + 2325) = 0.0350 and q - bar = 1 p - bar = 0.9650 alpha = 1%, critical value z = 2.326 B.

Test statistic and p-value. z(0.0466 - 0.0245) = 0.0221 / sqrt [(0.035 * 0.965 / 2081) + (0.035 * 0.965 / 2325)] 3.9849, p-value - 0.0000338 deliver the results at alpha = .01 Where the p-value is less than alpha you should reject Ho; there is significant evidence that the us e of this medicine reduces the result of de! aths. C. Is normality aware? 2081 * 0.035 = 72.84 > 5 and 2081 * 0.965 > 5 2325 * 0.035 = 70.86 > 5 and 2325 * 0.965 > 5 D. Is the residue large enough to be important? p-value has strong evidence that Ha is true. E. What else would medical researchers lack to know originally prescribing this drug widely? It could be the difference in family level and...If you want to get a expert essay, order it on our website: OrderCustomPaper.com

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